s/join/product/ ?
In APL's inner product (X a.b Y), each item of the result corresponds to a
row of X combined in some way with a column of Y. So the shape of the
result is (¯1↓⍴X),(1↓⍴Y).
In APL's outer product (X ∘.b Y), each item of the result corresponds to an
item of X combined in some way with an item of Y. So the shape of the
result is (⍴X),(⍴Y).
J probably does this in a more generic way, and you can certainly implement
inner and outer products and variations thereof by using the Rank operator.
For example:
X (b⍤99⍤0 99) Y ⍝ implements X ∘.b Y (where 99 is an approximation to
infinity)
X({+⌿⍺(×⍤¯1)⍵}⍤1 99)Y ⍝ implements X +.× Y
Jay.
On Tue, 14 Apr 2020 at 05:07, Elias Mårtenson <loke...@gmail.com> wrote:
> I was reading the spec on outer join to see if I could finally fully
> understand how it works.
>
> For reference, the spec refers to the syntax as follows:
>
> X a.b Y
>
> The function a is referenced twice in the description for this function,
> both times in the form a/ (i.e. with the reduce operator applied).
>
> I then noted that the J version of this function does not apply the reduce
> operator. I.e. in J, you have to do:
>
> X a/.b Y
>
> Now, am I correct in my assumption that with this change, there would be
> no need for the ∘ (null function)? I.e. if APL behaved like J, would the
> following achieve an outer join?
>
> X ⊣.× Y
>
> I'm still not sure I fully understand the join operator to the point where
> I can answer this question.
>
> Regards,
> Elias
>
